How do you show that $({sec}^{2} x + {csc}^{2} x) ({sec}^{2} x - {csc}^{2} x) = {tan}^{2} x - {cot}^{2} x$ $(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x$ ?

Question

How do you show that $({sec}^{2} x + {csc}^{2} x) ({sec}^{2} x - {csc}^{2} x) = {tan}^{2} x - {cot}^{2} x$ $(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x$ ?

1 Answer

Impact of this question

1714 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T17:00:29

The identity is false.

Explanation:

$({sec}^{2} x + {csc}^{2} x) ({sec}^{2} x - {csc}^{2} x) = {tan}^{2} x - {cot}^{2} x$ $(sec^2x + csc^2x)(sec^2x - csc^2x) = tan^2x - cot^2x$

$(\frac{1}{{cos}^{2} x} + \frac{1}{{sin}^{2} x}) (\frac{1}{{cos}^{2} x} - \frac{1}{{sin}^{2} x}) = \frac{{sin}^{2} x}{{cos}^{2} x} - \frac{{cos}^{2} x}{{sin}^{2} x}$ $(1/cos^2x + 1/sin^2x)(1/cos^2x- 1/sin^2x) = sin^2x/cos^2x - cos^2x/sin^2x$

$(\frac{{sin}^{2} x + {cos}^{2} x}{{cos}^{2} x {sin}^{2} x}) (\frac{{sin}^{2} x - {cos}^{2} x}{{cos}^{2} x {sin}^{2} x}) = \frac{{sin}^{2} x}{{cos}^{2} x} - \frac{{cos}^{2} x}{{sin}^{2} x}$ $((sin^2x + cos^2x)/(cos^2xsin^2x))((sin^2x - cos^2x)/(cos^2xsin^2x)) = sin^2x/cos^2x - cos^2x/sin^2x$

$\frac{{sin}^{2} x - {cos}^{2} x}{{cos}^{4} x {sin}^{4} x} = \frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x {cos}^{2} x}$ $(sin^2x- cos^2x)/(cos^4xsin^4x) = (sin^4x - cos^4x)/(sin^2xcos^2x)$

$\frac{{sin}^{2} x - {cos}^{2} x}{{cos}^{4} x {sin}^{4} x} = (({sin}^{2} x + {cos}^{2} x) \frac{{sin}^{2} x - {cos}^{2} x}{{sin}^{2} x {cos}^{2} x}$ $(sin^2x -cos^2x)/(cos^4xsin^4x) = ((sin^2x+ cos^2x)(sin^2x - cos^2x)/(sin^2xcos^2x)$

$\frac{{sin}^{2} x - {cos}^{2} x}{{cos}^{4} x {sin}^{4} x} = \frac{{sin}^{2} x - {cos}^{2} x}{{sin}^{2} x {cos}^{2} x}$ $(sin^2x - cos^2x)/(cos^4xsin^4x) = (sin^2x - cos^2x)/(sin^2xcos^2x)$

As you can see, the two sides are unequal, so the identity is false.

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you show that $({sec}^{2} x + {csc}^{2} x) ({sec}^{2} x - {csc}^{2} x) = {tan}^{2} x - {cot}^{2} x$ $(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you show that (sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x(sec2x+csc2x)(sec2x−csc2x)=tan2x−cot2x(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x?

1 Answer

Explanation:

Related questions

Impact of this question

How do you show that $({sec}^{2} x + {csc}^{2} x) ({sec}^{2} x - {csc}^{2} x) = {tan}^{2} x - {cot}^{2} x$ $(sec^2x + csc^2x)(sec^2x- csc^2x) =tan^2x -cot^2x$ ?